If "$i^2 = -1$" is an imaginary number, $i$ why is there no imaginary number for "$|x| = -1$"? The equation "$i^2 = -1$" has no real solution, so there's an entire number system of imaginary numbers that satisfy this.
So why is there not imaginary numbers that satisfy "$|x| = -1$"? There aren't any numbers, real or imaginary, that satisfy that equation, so shouldn't that make a new system of imaginary numbers?
 A: There are many problems if we define a number to be solution of this equation (some of which are already given in comments above):

*

*Let $x$ be a number such that $|x|=-1$. Then $x^2=1\implies(x-1)(x+1)=0$. Since both the terms on LHS are non-zero, the number system you're defining isn't even an integral domain.


*$i^2=-1$ makes sense: since $-1$ shows one step backward on number line ($180^\circ$ rotation), and $i$ shows $90^\circ$ rotation in argand plane (which is essentially an extension of number line in 2D). There is no significance to the one you're suggesting. A point can't have negative distance from the origin.


*Again as suggested in comments, it won't be  a metric then.
A: $|x|\,$ is always a non-negative real number even though $\,x\,$ is a complex number, hence $\,|x|=-1\,$ cannot have any solution.
So far there does not exist any number which satisfies $\,|x|=-1\,$ and, what is more, if there existed such a number, it would not be an element of a field.
Moreover, if there existed such a number, then the following properties
$1)\quad |ab|=|a||b|\;\;,$
$2)\quad |a+b|\leqslant |a|+|b|\;\;,$
would be no longer valid, indeed if they were correct, we would get that
$|-x|=|(-1)\cdot x|=|-1||x|=1|x|=|x|\;\;,$
$0=|0|=|x+(-x)|\leqslant|x|+|-x|=2|x|=-2\;\;,$
but it is a contradiction.
